\(\int \frac {e^{\frac {-2-x}{(-x+2 x^3) \log (x)}} (-2-x+4 x^2+2 x^3+(-2+12 x^2+4 x^3) \log (x))}{(x^2-4 x^4+4 x^6) \log ^2(x)} \, dx\) [2019]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-2)]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 74, antiderivative size = 19 \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=e^{\frac {2+x}{\left (x-2 x^3\right ) \log (x)}} \]

[Out]

exp((2+x)/ln(x)/(-2*x^3+x))

Rubi [F]

\[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=\int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^((-2 - x)/((-x + 2*x^3)*Log[x]))*(-2 - x + 4*x^2 + 2*x^3 + (-2 + 12*x^2 + 4*x^3)*Log[x]))/((x^2 - 4*x^4
 + 4*x^6)*Log[x]^2),x]

[Out]

-2*Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/(x^2*Log[x]^2), x] - Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)
*Log[x]))/(x*Log[x]^2), x] - 2*Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/((1 - Sqrt[2]*x)*Log[x]^2), x]
- Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/((1 - Sqrt[2]*x)*Log[x]^2), x]/Sqrt[2] - 2*Defer[Int][E^((-2
 - x)/(x*(-1 + 2*x^2)*Log[x]))/((1 + Sqrt[2]*x)*Log[x]^2), x] + Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x])
)/((1 + Sqrt[2]*x)*Log[x]^2), x]/Sqrt[2] - 2*Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/(x^2*Log[x]), x]
- 2*Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/((1 - Sqrt[2]*x)*Log[x]), x] - 2*Defer[Int][E^((-2 - x)/(x
*(-1 + 2*x^2)*Log[x]))/((1 + Sqrt[2]*x)*Log[x]), x] + 8*Defer[Int][E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))/((-1 +
 2*x^2)^2*Log[x]), x] + 4*Defer[Int][(E^((-2 - x)/(x*(-1 + 2*x^2)*Log[x]))*x)/((-1 + 2*x^2)^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{x^2 \left (1-4 x^2+4 x^4\right ) \log ^2(x)} \, dx \\ & = 4 \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{x^2 \left (-2+4 x^2\right )^2 \log ^2(x)} \, dx \\ & = 4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{x^2 \left (2-4 x^2\right )^2 \log ^2(x)} \, dx \\ & = 4 \int \left (\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{4 x^2 \left (-1+2 x^2\right ) \log ^2(x)}+\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} \left (-1+6 x^2+2 x^3\right )}{2 x^2 \left (-1+2 x^2\right )^2 \log (x)}\right ) \, dx \\ & = 2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} \left (-1+6 x^2+2 x^3\right )}{x^2 \left (-1+2 x^2\right )^2 \log (x)} \, dx+\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{x^2 \left (-1+2 x^2\right ) \log ^2(x)} \, dx \\ & = 2 \int \left (-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)}+\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{\left (-1+2 x^2\right )^2 \log (x)}+\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right ) \log (x)}\right ) \, dx+\int \left (-\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)}-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)}+\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{\left (-1+2 x^2\right ) \log ^2(x)}\right ) \, dx \\ & = -\left (2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)} \, dx\right )+2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{\left (-1+2 x^2\right ) \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)} \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} (2+x)}{\left (-1+2 x^2\right )^2 \log (x)} \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right ) \log (x)} \, dx-\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)} \, dx \\ & = 2 \int \left (\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right ) \log ^2(x)}+\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right ) \log ^2(x)}\right ) \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)} \, dx+4 \int \left (-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \left (1-\sqrt {2} x\right ) \log (x)}-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \left (1+\sqrt {2} x\right ) \log (x)}\right ) \, dx+4 \int \left (\frac {2 e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right )^2 \log (x)}+\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right )^2 \log (x)}\right ) \, dx-\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)} \, dx \\ & = -\left (2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)} \, dx\right )+2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right ) \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1-\sqrt {2} x\right ) \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1+\sqrt {2} x\right ) \log (x)} \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right ) \log ^2(x)} \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right )^2 \log (x)} \, dx+8 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right )^2 \log (x)} \, dx-\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)} \, dx \\ & = 2 \int \left (-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \sqrt {2} \left (1-\sqrt {2} x\right ) \log ^2(x)}+\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \sqrt {2} \left (1+\sqrt {2} x\right ) \log ^2(x)}\right ) \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1-\sqrt {2} x\right ) \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1+\sqrt {2} x\right ) \log (x)} \, dx+4 \int \left (-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \left (1-\sqrt {2} x\right ) \log ^2(x)}-\frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{2 \left (1+\sqrt {2} x\right ) \log ^2(x)}\right ) \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right )^2 \log (x)} \, dx+8 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right )^2 \log (x)} \, dx-\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)} \, dx \\ & = -\left (2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log ^2(x)} \, dx\right )-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1-\sqrt {2} x\right ) \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1+\sqrt {2} x\right ) \log ^2(x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x^2 \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1-\sqrt {2} x\right ) \log (x)} \, dx-2 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1+\sqrt {2} x\right ) \log (x)} \, dx+4 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}} x}{\left (-1+2 x^2\right )^2 \log (x)} \, dx+8 \int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (-1+2 x^2\right )^2 \log (x)} \, dx-\frac {\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1-\sqrt {2} x\right ) \log ^2(x)} \, dx}{\sqrt {2}}+\frac {\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{\left (1+\sqrt {2} x\right ) \log ^2(x)} \, dx}{\sqrt {2}}-\int \frac {e^{\frac {-2-x}{x \left (-1+2 x^2\right ) \log (x)}}}{x \log ^2(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.78 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=e^{\frac {2+x}{x \log (x)-2 x^3 \log (x)}} \]

[In]

Integrate[(E^((-2 - x)/((-x + 2*x^3)*Log[x]))*(-2 - x + 4*x^2 + 2*x^3 + (-2 + 12*x^2 + 4*x^3)*Log[x]))/((x^2 -
 4*x^4 + 4*x^6)*Log[x]^2),x]

[Out]

E^((2 + x)/(x*Log[x] - 2*x^3*Log[x]))

Maple [A] (verified)

Time = 4.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21

method result size
risch \({\mathrm e}^{-\frac {2+x}{x \left (2 x^{2}-1\right ) \ln \left (x \right )}}\) \(23\)
parallelrisch \({\mathrm e}^{-\frac {2+x}{x \left (2 x^{2}-1\right ) \ln \left (x \right )}}\) \(23\)

[In]

int(((4*x^3+12*x^2-2)*ln(x)+2*x^3+4*x^2-x-2)*exp((-2-x)/(2*x^3-x)/ln(x))/(4*x^6-4*x^4+x^2)/ln(x)^2,x,method=_R
ETURNVERBOSE)

[Out]

exp(-(2+x)/x/(2*x^2-1)/ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=e^{\left (-\frac {x + 2}{{\left (2 \, x^{3} - x\right )} \log \left (x\right )}\right )} \]

[In]

integrate(((4*x^3+12*x^2-2)*log(x)+2*x^3+4*x^2-x-2)*exp((-2-x)/(2*x^3-x)/log(x))/(4*x^6-4*x^4+x^2)/log(x)^2,x,
 algorithm="fricas")

[Out]

e^(-(x + 2)/((2*x^3 - x)*log(x)))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((4*x**3+12*x**2-2)*ln(x)+2*x**3+4*x**2-x-2)*exp((-2-x)/(2*x**3-x)/ln(x))/(4*x**6-4*x**4+x**2)/ln(x)
**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(((4*x^3+12*x^2-2)*log(x)+2*x^3+4*x^2-x-2)*exp((-2-x)/(2*x^3-x)/log(x))/(4*x^6-4*x^4+x^2)/log(x)^2,x,
 algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
 of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95 \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx=e^{\left (-\frac {x}{2 \, x^{3} \log \left (x\right ) - x \log \left (x\right )} - \frac {2}{2 \, x^{3} \log \left (x\right ) - x \log \left (x\right )}\right )} \]

[In]

integrate(((4*x^3+12*x^2-2)*log(x)+2*x^3+4*x^2-x-2)*exp((-2-x)/(2*x^3-x)/log(x))/(4*x^6-4*x^4+x^2)/log(x)^2,x,
 algorithm="giac")

[Out]

e^(-x/(2*x^3*log(x) - x*log(x)) - 2/(2*x^3*log(x) - x*log(x)))

Mupad [B] (verification not implemented)

Time = 8.97 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\frac {-2-x}{\left (-x+2 x^3\right ) \log (x)}} \left (-2-x+4 x^2+2 x^3+\left (-2+12 x^2+4 x^3\right ) \log (x)\right )}{\left (x^2-4 x^4+4 x^6\right ) \log ^2(x)} \, dx={\mathrm {e}}^{\frac {1}{\ln \left (x\right )-2\,x^2\,\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {2}{2\,x^3\,\ln \left (x\right )-x\,\ln \left (x\right )}} \]

[In]

int((exp((x + 2)/(log(x)*(x - 2*x^3)))*(log(x)*(12*x^2 + 4*x^3 - 2) - x + 4*x^2 + 2*x^3 - 2))/(log(x)^2*(x^2 -
 4*x^4 + 4*x^6)),x)

[Out]

exp(1/(log(x) - 2*x^2*log(x)))*exp(-2/(2*x^3*log(x) - x*log(x)))